Vtyjkyuk

Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite Let $A in M_n(mathbbC)$ be a hermitian matrix that satisfies the property $A^99=I_n$. I want to show that $A=I_n$. By definition, $A$ is a hermitian matrix if $a_ij=overlinea_ji$. From the fact that $A^99=I_n$, we deduce that $A$ is invertible and its inverse is $A^98$. Does that help to get that $A=I_n$ ? Or do we use somehow the definition of a hermitian matrix? linear-algebra matrices share | cite | improve this question edited Sep 2 at 12:42 Shaun 7,447 9 29 72 asked Sep 2 at 12:39 Evinda 552 4 12 3 Hint: Since $A$ is hermitian, it can be diagonalized. It is easy to write down the entries of a power of a diagonal matrix. And remember that the eigenvalues of $A$ are real. – Henning Makholm Sep 2 at 12:42 So since $A$ is hermitian, there exists an invertible matrix $P$ such that $P^-1AP$ is a diagonal matrix. How does this help? @HenningMakho

Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite Problem definition : Show that image of $S$ under $A$, i.e., $A(S) = Ax : x in S $ is (a) convex and (b) but not closed in general if $S subseteq mathbbR^n$ is convex. Also, can we show when $A(S)$ is closed and when it is not? However, if we consider the preimage $S$ under $A$, i.e., $A^-1(S) = x : Ax in S $, then it is both closed and convex. EDIT: $S subseteq mathbbR^n$ is convex and $A in mathbbR^m times n$. EDIT EDIT: Additionally, assume that $S$ is closed, i.e., $S$ is a closed and convex set. real-analysis convex-analysis share | cite | improve this question edited Sep 4 at 5:01 asked Sep 2 at 12:45 user550103 721 2 14 $A$ is a linear transformation here? Also, it seems like the exercise should read “$A(S)$ isn’t closed in general ”. Sometimes $A(S)$ is closed, but it’s not always closed. – David M. Sep 2 at 14:31